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julsineya
9 days ago
14

15. How many grams are equal to 0.11 mole of copper(1) chromate, Cu2(CrO4)? with work please​

Chemistry
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The dipole moment (μ) of hbr (a polar covalent molecule) is 0.797d (debye), and its percent ionic character is 11.8 % . estimate
Alekssandra [3086]
The dipole moment u can be calculated using the formula
U = rq
Where u represents the dipole moment
R indicates the bond length
Q = 1.6x10-19 C  
Hence, R = u/q
R = (0.797 d) ( 3.34x10^-30 Cm/ 1 d) /( 1.6x10^-19 C)(0.118)
R = 1.41x10^-10 m
<span>R = 141 pm</span>
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How would a flood be a limitation for synthetic polymers that rely on natural rubber in its production?
eduard [2782]
A flood that affects the environment where natural rubber is produced would severely hinder rubber production. In order to greatly limit production, a flood would need to destroy a significant portion of rubber trees. Natural rubber is crucial for manufacturing synthetic polymers. If the rubber supply is compromised (due to the disruption of its ecosystem caused by a flood), there would be a substantial decline in the availability of synthetic polymers.


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In the first order decomposition of acetone at 500°c, ch3och3→ productit is found that the concentration of acetone is 0.0300 m
lorasvet [2795]
For the first-order decomposition, the equation is: ln(x0 / x) = kt. At t = 200, x = 0.0300 M, we have ln(x0 / 0.03) = 200k. At t = 400, when x = 0.0200 M, we utilize ln(x0 / 0.02) = 400k. By multiplying the first equation by 2, we get 2ln(x0 / 0.03) = 400k, which aligns with the second equation, leading us to conclude that 2ln(x0 / 0.03) = ln(x0 / 0.02). This suggests (x0 / 0.03)^2 = x0 / 0.02, allowing us to find x0 = 0.045 M as the initial concentration. Plugging this back into the first equation yields: ln(0.045 / 0.03) = 200k, from which it follows that k = 0.0020273 (rate constant). The half-life can be calculated with x = 0.5x0: ln(x0 / 0.5x0) = 0.0020273t, resulting in ln(2) = 0.0020273t, which simplifies to t = 341.90 minutes (half-life).
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